Introduction Construction of Lax pairs Conclusions A Symbolic Algorithm to Compute Lax Pairs in Matrix Form for Nonlinear Evolution Equations Jacob Rezac1 Willy Hereman2 1 University of Delaware School of Mines IMACS Nonlinear Evolution Equations and Wave Phenomena: Computation and Theory 2013 2 Colorado 03/26/2013 J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions 1 Introduction 2 Construction of Lax pairs 3 Conclusions Collaborators: Sara Clifton, Oscar Aguilar (CSM) Funded by NSF research award no. CCF-0830783 J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions What is a matrix Lax pair? For a PDE, ut = f (u, ux , uxx , ...), matrices X and T which satisfy Φx = X Φ and Φt = T Φ, for an eigenfunction Φ are called a Lax pair. This is equivalent to the equation Xt − Tx + [X , T ] = ˙ 0 for [X , T ] = XT − TX . The notation = ˙ means equality on solutions of the PDE. J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Example Consider the Korteweg-de Vries (KdV) equation, ut + αuux + uxxx = 0. A known Lax pair is XKdV = TKdV = −4λ2 + 0 1 λ − 16 αu 0 1 6 αux 1 1 2 2 αλu + 18 α u 3 , + 16 αuxx −4λ − 13 αu − 61 αux . The isospectral parameter λ (λt = 0) is necessary for non-triviality. J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Example Substituting these into the Lax equation, 0 0 (XKdV )t − (TKdV )x + [XKdV , TKdV ] = − 16 α (ut + αuux + uxxx ) 0 0 0 = ˙ . 0 0 J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Operators We can also use operators, L and M which satisfy Lt + [L, M] = ˙ 0. We don’t consider these here. M. Hickman, W. Hereman, J. Larue, Ü. Göktas, Scaling invariant Lax pairs of nonlinear evolution equations, Applicable Analysis. Vol. 91 (2), pp. 381-402 (2012). J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Applications Lax pairs are useful in Finding exact solutions of nonlinear PDEs (inverse scattering transform) Construction of conservation laws (see my poster tomorrow!) “Verification” that an equation is integrable. J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions How could we construct this? Try “linear” elements for X X1,1 + X2,1 λ + X3,1 u X1,2 + X2,2 λ + X3,2 u XGuess = X1,3 + X2,3 λ + X3,3 u X1,4 + X2,4 λ + X3,4 u and up to “quadratic” elements for T , e.g., (1,1) TGuess = T1,1 + T2,1 λ + T3,1 u + T4,1 ux + T5,1 λ2 + T6,1 λu + T7,1 u 2 + T8,1 uxx . J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions How could we construct this? Substituting into the Lax equation, XGuess,t − TGuess,x L1,1 L2,2 + [XGuess , TGuess ] = , L2,1 L2,2 where, e.g., L1,1 = −λT6,1 ux + λT4,3 X2,2 ux − λT4,2 X2,3 ux + λT8,3 X2,2 uxx − λT8,2 X2,3 uxx + T4,3 X1,2 ux − T4,2 X1,3 ux + T4,3 X3,2 uux − T4,2 X3,3 uux + T8,3 X1,2 uxx − T8,2 X1,3 uxx + T8,3 X3,2 uuxx − T8,2 X3,3 uuxx − T3,1 ux − 2T7,1 uux − T4,1 uxx − T8,1 uxxx + λ2 T6,3 X2,2 u − λ2 T6,2 X2,3 u + λ2 T5,3 X3,2 u − λ2 T5,2 X3,3 u + λT6,2 X1,3 u + λT7,3 X2,2 u 2 + λT3,3 X2,2 u − λT7,2 X2,3 u 2 − . . . J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions How could we construct this? . . . − λT3,2 X2,3 u + λT6,3 X3,2 u 2 + λT2,3 X3,2 u − λT6,2 X3,3 u 2 − λT2,2 X3,3 u + T7,3 X1,2 u 2 + T3,3 X1,2 u − T7,2 X1,3 u 2 − T3,2 X1,3 u+ T7,3 X3,2 u 3 + T3,3 X3,2 u 2 + T1,3 X3,2 u − T7,2 X3,3 u 3 − T3,2 X3,3 u 2 − T1,2 X3,3 u − αX3,1 uux − X3,1 uxxx + λ3 T5,3 X2,2 − λT6,3 X1,2 u− T3,2 X1,3 u − λT6,2 X1,3 u + λT3,3 X2,2 u + λ3 T5,2 X2,3 + λ2 T5,3 X1,2 − λ2 T5,2 X1,3 + λ2 T2,3 X2,2 − λ2 T2,2 X2,3 + λT2,3 X1,2 − λT2,2 X1,3 + λT1,3 X2,2 − λT1,2 X2,3 + T1,3 X1,2 − T1,2 X1,3 . J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions How could we construct this? This is pretty bad! We can clean this up by matching powers. For example, L1,1 = λ (T2,3 X1,2 − T2,2 X1,3 + T1,3 X2,2 − T1,2 X2,3 ) + other terms. It is simpler to solve T2,3 X1,2 − T2,2 X1,3 + T1,3 X2,2 − T1,2 X2,3 = 0, especially with the other equations from L1,1 , L1,2 , L2,1 , and L2,2 . J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Big Idea Use properties of PDE to simplify guess. In particular, consider scaling invariance. J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Scaling Invariance and Weight Rescale the variables in the KdV equation to (x, t, u) → (κ−1 x, κ−3 t, κ2 u) = (ξ, τ, ν). (1) Then, ut + αuux + uxxx = 0 → κ−5 (ντ + αννξ + νξξξ ) = 0. Call each exponent on κ in (1) the weight of a variable (denoted W (·)). That is, ∂ ∂ W =1 W =3 W (u) = 2. ∂x ∂t J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Consider again the KdV equation X matrix, weight 0 weight 0 z}|{ z}|{ 0 1 1 XKdV = λ − αu 0 |{z} | {z6 } weight 0 weight 2 We have W ∂ ∂x =1 W J. Rezac ∂ ∂t =3 W (u) = 2. Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Consider again the KdV equation X matrix, weight 0 weight 0 z}|{ z}|{ 0 1 1 XKdV = λ − αu 0 |{z} | {z6 } weight 0 weight 2 We have W ∂ ∂x =1 W J. Rezac ∂ ∂t =3 W (u) = 2. Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Consider again the KdV equation X matrix, weight 0 weight 0 z}|{ z}|{ 0 1 1 XKdV = λ − αu 0 |{z} | {z6 } weight 0 weight 2 We have W ∂ ∂x =1 W J. Rezac ∂ ∂t =3 W (u) = 2. Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Consider again the KdV equation X matrix, weight 0 weight 0 z}|{ z}|{ 0 1 1 XKdV = λ − αu 0 |{z} | {z6 } weight 0 weight 2 We have W ∂ ∂x =1 W J. Rezac ∂ ∂t =3 W (u) = 2. Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Consider again the KdV equation X matrix, weight 0 weight 0 z}|{ z}|{ 0 1 1 XKdV = λ − αu 0 |{z} | {z6 } weight 0 weight 2 We have W ∂ ∂x =1 W J. Rezac ∂ ∂t =3 W (u) = 2. Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Consider again the KdV equation X matrix, weight 0 weight 0 z}|{ z}|{ 0 1 1 XKdV = λ − αu 0 |{z} | {z6 } weight 0 weight 2 We have W ∂ ∂x =1 W ∂ ∂t =3 W (u) = 2. Why is W (λ) = 2? Set λ = iκ2 . (We will implicitly take spectral parameter weight to be 1). J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions The modified KdV (mKdV) equation, ut + αu 2 ux + uxxx = 0, with weights W ∂ ∂x =1 W ∂ ∂t =3 has Lax pair with T -component weight 3 z }| { −4iλ3 − 2iλu 2 TmKdV = 4λ2 u − 2iλux + 2u 3 − uxx | {z } weight 3 J. Rezac W (u) = 1 weight 3 z }| { 4λ2 u + 2iλux + 2u 3 − uxx 3 2 |4iλ + {z2iλu} Computation of Lax Pairs weight 3 Introduction Construction of Lax pairs Conclusions The modified KdV (mKdV) equation, ut + αu 2 ux + uxxx = 0, with weights W ∂ ∂x =1 W ∂ ∂t =3 has Lax pair with T -component weight 3 z }| { −4iλ3 − 2iλu 2 TmKdV = 4λ2 u − 2iλux + 2u 3 − uxx | {z } weight 3 J. Rezac W (u) = 1 weight 3 z }| { 4λ2 u + 2iλux + 2u 3 − uxx 3 2 |4iλ + {z2iλu} Computation of Lax Pairs weight 3 Introduction Construction of Lax pairs Conclusions The modified KdV (mKdV) equation, ut + αu 2 ux + uxxx = 0, with weights W ∂ ∂x =1 W ∂ ∂t =3 has Lax pair with T -component weight 3 z }| { −4iλ3 − 2iλu 2 TmKdV = 4λ2 u − 2iλux + 2u 3 − uxx | {z } weight 3 J. Rezac W (u) = 1 weight 3 z }| { 4λ2 u + 2iλux + 2u 3 − uxx 3 2 |4iλ + {z2iλu} Computation of Lax Pairs weight 3 Introduction Construction of Lax pairs Conclusions The modified KdV (mKdV) equation, ut + αu 2 ux + uxxx = 0, with weights W ∂ ∂x =1 W ∂ ∂t =3 has Lax pair with T -component weight 3 z }| { −4iλ3 − 2iλu 2 TmKdV = 4λ2 u − 2iλux + 2u 3 − uxx | {z } weight 3 J. Rezac W (u) = 1 weight 3 z }| { 4λ2 u + 2iλux + 2u 3 − uxx 3 2 |4iλ + {z2iλu} Computation of Lax Pairs weight 3 Introduction Construction of Lax pairs Conclusions The modified KdV (mKdV) equation, ut + αu 2 ux + uxxx = 0, with weights W ∂ ∂x =1 W ∂ ∂t =3 has Lax pair with T -component weight 3 z }| { −4iλ3 − 2iλu 2 TmKdV = 4λ2 u − 2iλux + 2u 3 − uxx | {z } weight 3 J. Rezac W (u) = 1 weight 3 z }| { 4λ2 u + 2iλux + 2u 3 − uxx 3 2 |4iλ + {z2iλu} Computation of Lax Pairs weight 3 Introduction Construction of Lax pairs Conclusions Construction Assume Lax matrices have weight invariant elements, e.g., W (X11 ) W (X12 ) W (T11 ) W (T12 ) W (X ) = W (T ) = . W (X21 ) W (X22 ) W (T21 ) W (T22 ) We substitute these weights into the Lax equation and force scale invariance, W (Xt ) = W (Tx ) = W ([X , T ]) , or, W (X ) + W ∂ ∂x J. Rezac ∂ = W (T ) + W ∂t = W (X ) + W (T ) . Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Solving the system of equations produced from this yields ∂ W ∂x W (X12) W (X ) = ∂ ∂ − W (X12 ) W ∂x 2W ∂x and W (T ) = ∂ W ∂t ∂ W ∂t − W (X12 ) + W J. Rezac ∂ ∂x W ∂ ∂t + W (X12) − W ∂ W ∂t Computation of Lax Pairs ∂ ∂x . Introduction Construction of Lax pairs Conclusions Method of Construction Step 1: Pick W (X12 ) J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Method of Construction Step 1: Pick W (X12 ) Step 2: Based on the weight matrices of X and T , generate elements of X and T with undetermined coefficients. J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Method of Construction Step 1: Pick W (X12 ) Step 2: Based on the weight matrices of X and T , generate elements of X and T with undetermined coefficients. Step 3: Substitute matrices into Lax equation. J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Method of Construction Step 1: Pick W (X12 ) Step 2: Based on the weight matrices of X and T , generate elements of X and T with undetermined coefficients. Step 3: Substitute matrices into Lax equation. Step 4: Solve for undetermined coefficients. J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Method of Construction Step 1: Pick W (X12 ) Step 2: Based on the weight matrices of X and T , generate elements of X and T with undetermined coefficients. Step 3: Substitute matrices into Lax equation. Step 4: Solve for undetermined coefficients. Step 5: Substitute back into X and T . J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Example Consider the KdV equation (α = 6) and assume (Step 1) W (X12 ) = 0. Then, our candidate Lax pair is (Step 2) c1 κ c3 X = c4 κ2 + c5 u −c1 κ and T = T11 T12 T21 −T11 , where T11 = c6 κ3 + c7 κu + c8 ux , T12 = c9 κ2 + c10 u, T21 = c11 κ4 + c12 κ2 u + c13 κux + c14 u 2 + c15 uxx . J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions We substitute this into the Lax equation (Step 3) and get, for the (1, 1)-element, (c11 c3 − c4 c9 ) κ4 + (c12 c3 − c10 c4 − c5 c9 ) κ2 u + (c14 c3 − c10 c5 ) u 2 + (c13 c3 − c7 ) κux + (c15 c3 − c8 ) uxx = 0. We break this up in powers of κ and u to get the system of equations c11 c3 − c4 c9 = 0, c12 c3 − c10 c4 − c5 c9 = 0, c14 c3 − c10 c5 = 0, c13 c3 − c7 = 0, c15 c3 − c8 = 0. J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Solving with the equations from the other three matrix elements (Step 4) gives many solutions. Choose (Step 5) a one: # " 0 − cc10 2 11 X = c11 2 + 2c u c κ 0 10 11 2c10 and T = T11 T12 T21 −T11 , where T11 = ux , 2c10 T12 = 3 −c10 κ2 + c11 u , c11 T21 = c10 κ4 + c11 κ2 u − 2 c11 2u 2 + uxx . c10 Setting c10 = 4, c11 = 2i, and κ2 = −iλ gives the pair from before. J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions What about W (X12 )? Assume X and T matrices are of the form X = n X X (i) λi and T = i=0 m X T (j) λj j=0 where weights are chosen such that W (λ) = 1 and each element of ∂ and each element of T has weight X has weight n = W ∂x ∂ m = W ∂t . That is, weight n z X (i) |{z} }| { × |{z} λi weight n − i J. Rezac weight i Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Substitute our series forms of X and T into the Lax equation. These can be separated by powers of λ and split into two types of ∂ equations. Assuming W ∂x = 1, Algebraic constraints [X (1) , T (m) ] = ˙ 0 and 1 X [X (k) , T (m−k) ] = ˙ 0 k=0 Solve these algebraic equations and substitute into Differential constraints (0) (0) − T + [X (0) , T (0) ] = ˙ 0 X t x − T (i) x + 1 X [X (k) , T (m−i−k) ] = ˙ 0 k=0 These can now be solved by imposing weights. J. Rezac Computation of Lax Pairs i = 1, . . . , m − 1 Introduction Construction of Lax pairs Conclusions Example Consider the mKdV equation and assume matrices of the form X = 1 X X (i) λi and i=0 T = 3 X T (j) λj . j=0 The algebraic constraints are [X (1) , T (3) ] = 0, [X (0) , T (3) ] + [X (1) , T (2) ] = 0. To solve this, we set up J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions " X11 X12 (i) (i) X21 −X11 " T11 T12 (i) (i) T21 −T11 X (i) = and T (i) = (i) (i) # (i) (i) # . Solving, we find J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions X11 X12 X21 −X11 X = , where (0) (1) X11 = X11 + X11 λ (3) X12 = (0) (1) (3) (1) (3) (2) (0) (2) (1) (3) (3) − T11 T21 X11 (3) (T11 )2 (2) T11 Substituting in candidate elements gives J. Rezac (1) T11 T12 X11 (1) T21 X11 + T21 X11 + T21 X11 λ (3) − T11 (3) X21 = (0) T12 X11 + T12 X11 + T12 X11 λ Computation of Lax Pairs (3) (3) (T11 )2 (1) . Introduction Construction of Lax pairs Conclusions X = X11 X12 X21 −X11 and T = T11 T12 T21 −T11 ., where X11 = c1 λ + c2 u 1 c1 c3 c5 X12 = c1 c7 λ + c2 c3 + c1 c6 − u c4 c4 c1 c5 c7 1 c1 c7 λ + c2 c7 + c1 c8 − u . X21 = c4 c4 and T11 = c4 λ3 + c5 uλ2 + (c11 u 2 + c12 ux )λ + c9 u 3 + c10 uux + c13 uxx , T12 = c3 λ3 + c6 uλ2 + (c16 u 2 + c17 ux )λ + c14 u 3 + c15 uux + c18 uxx T21 = c7 λ3 + c8 uλ2 + (c21 u 2 + c22 ux )λ + c19 u 3 + c20 uux + c23 uxx . We use these to solve the constraints with derivatives. Doing so J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions c1 λ − c22 u 2c2 u −c1 λ T11 T12 T21 −T11 X = and T = . where T11 = −4c13 λ3 − 2c1 λu 2 , 1 T12 = 2c12 c2 λ2 u + c1 c2 λux + c2 u 3 + c2 uxx 2 2 T21 = −4c1 λ2 u + 2c1 λux − 2u 3 − uxx . c2 Setting c1 = i and c2 = 2 gives the same answer as before. J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Problems with Method Large number of possible values for undetermined coefficients. J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Problems with Method Large number of possible values for undetermined coefficients. We can only handle simple forms for Lax pairs. J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Problem 1: Short Pulse Equation The Short Pulse Equation (SPE), uxt + αu + β(u 3 )xx = 0, with weights ∂ W =1 ∂x W ∂ ∂t =3 W (u) = 1 has a Lax pair with X -component λ XSPE = 1 α λux λux λ W (α) = 4 This can be constructed with the method proposed in this talk. At present, we wouldn’t if we didn’t already know the answer. J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Problems with Method Large number of possible values for undetermined coefficients. We can only handle simple forms for Lax pairs. Gauge freedom X̂ = GXG −1 + Gx G −1 J. Rezac and T̂ = GTG −1 + Gt G −1 Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Problem 2: Weak Lax Pairs In, e.g., S.Y. Sakovich. On zero-curvature representations of evolution equations. Journal of Physics A: Mathematical and General, 28(10):28612869, 1995., the concept of a weak Lax pair is discussed. Burgers’ equation, ut − uxx − uux = 0 can be expressed with the Lax pair 1 λ λ + 12 u X = λ 2 −λ + 12 u and 1 λu T = 4 −λu + 21 u 2 + u + x J. Rezac λu + 12 u 2 + ux . −λu Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Problem 2: Weak Lax Pairs However, under the gauge transformation "√ √ λ + √1λ λ− 1 √ √ G= 1 λ − √λ λ+ 2 we have √1 λ √1 λ # , 1 1 1 + 12 u X̂ = −1 2 −1 + 12 u and 1 T̂ = 4 u 1 2 2 u − u + ux 1 2 2u + u + ux . −u Free of λ! J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Problems with Method Large number of possible values for undetermined coefficients. We can only handle simple forms for Lax pairs. Gauge freedom X̂ = GXG −1 + Gx G −1 J. Rezac and T̂ = GTG −1 + Gt G −1 Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Problems with Method Large number of possible values for undetermined coefficients. We can only handle simple forms for Lax pairs. Gauge freedom X̂ = GXG −1 + Gx G −1 and T̂ = GTG −1 + Gt G −1 Use is limited to scaling invariant PDEs J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Problems with Method Large number of possible values for undetermined coefficients. We can only handle simple forms for Lax pairs. Gauge freedom X̂ = GXG −1 + Gx G −1 and T̂ = GTG −1 + Gt G −1 Use is limited to scaling invariant PDEs Computational complexity J. Rezac Computation of Lax Pairs Introduction Construction of Lax pairs Conclusions Conclusions Thanks! J. Rezac Computation of Lax Pairs